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Let $(X_n)_nge0$ be a markov chain on $0,1,...$ with transition probabilities given by

$p_01=1,p_i,i+1+p_i,i-1=1, p_i,i+1=Big(fraci+1iBig)^2p_i,i-1, ige1$

I need to show that if $X_0=0$ then the probability that $X_nge1$ for all $nge 1$ is $frac6pi^2$

I'm looking for tips on how to solve this task.

Here is a rather long answer but it has the merit of being quite general I think.

Minimality of the hitting probability

In the general case of a Markov chain with state space $mathbbN$ and transitions $(p_i,j)_(i,j)inmathbbN^2$, consider $mathcalP subset mathbbN$ a subset of states and for all $i inmathbbN$, $(X_n(i))$ a Markov chain starting from $i$ and the hitting probability $h_mathcalP(i) = mathbbPbig(infnge 0, X_n(i) in mathcalP < +inftybig)$. Now we have that $$h_mathcalP mbox is the minimal non negative solution to quad h(i)=left{beginarrayll 1 mbox if i in mathcalP\ sum limits_j in mathbbN p_i,jh(j) mbox otherwise.endarrayright.$$

Indeed, $h_mathcalP$ is a solution. Conversely, if $h$ is a solution, for $i in mathbbN backslash mathcalP$, $h(i) = sum limits_j in mathcalP p_i,j + sum limits_j in mathbbNbackslash mathcalP p_i,jh(j)$. Plugging it back, we also get $h(i) = sum limits_j in mathcalP p_i,j + sum limits_j in mathbbNbackslashmathcalPp_i,jsum limits_k in mathcalPp_j,k + sum limits_j in mathbbNbackslashmathcalPp_i,j sum limits_k in mathbbNbackslash mathcalP p_j,kh(k)$. Repeat several times and note that the last term is non negative: you get $$h(i) ge mathbbP(X_1 in mathcalP)+mathbbP(X_1 notin mathcalP,X_2inmathcalP)+cdotcdotcdot+mathbbP(X_1notinmathcalP,...,X_n-1notinmathcalP,X_ninmathcalP).$$ Taking the limit, $h(i) ge h_mathcalP(i)$ using the definition of $h_mathcalP$.

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Back to your problem

I will change a bit your problem. We are going to see $0$ as a well, and the random walk starts from $1$. So $X_0=1$ and $p_0,0 = 1$. Nothing else is changed.

In your problem the hitting states are just $mathcalP = 0$, the transition are given as you stated. Let us define and compute the sequence $u_n = h_mathcalP(n)$, the probability of hitting $0$ starting from $n$. We have $u_0 = 1$ (if you start in the well $0$, you hit it) and for all $n ge 1$, the law of total probability yields $u_n = p_n,n-1u_n-1+p_n,n+1u_n+1 = fracn^2n^2+(n+1)^2u_n-1+frac(n+1)^2n^2+(n+1)^2u_n+1$. Regrouping gives you $(n+1)^2 (u_n+1-u_n) = n^2 (u_n-u_n-1)$ so $u_n+1-u_n = Big(prod limits_k=1^n frack^2(k+1)^2Big)(u_1-u_0)=frac1(n+1)^2(u_1-1)$.

Summing, we get $u_n-u_0 = sum limits_k=0^n-1 frac1(k+1)^2(u_1-1)$ i.e. $$u_n = 1 - (1-u_1) sum limits_k=1^n frac1k^2.$$

Now use the first section: among all possible choices of $u_1$, it has to be the one that makes $(u_n)$ minimal and non negative. Remind that $sum limits_k=1^n frac1k^2 undersetn to inftylongrightarrow fracpi^26$. If we had $u_1 < 1-frac6pi^2$, $(u_n)$ would not be non-negative, and with $u_1 > 1-frac6pi^2$ it would not be the minimal solution (because a $u_1'$ between $1-frac6pi^2$ and $u_1$ would give a smaller solution).

That gives you $u_1 = 1-frac6pi^2$. The probability of going back to zero starting from $1$ is $1-frac6pi^2$. Additionnally, we found that the probabilty of going back to zero starting from $n$ is $1 - frac6pi^2 sum limits_k=1^n frac1k^2$.

Hint 1: it's easier to find the probability of the complement event.

Hint 2: you can use the first entrance decomposition to write
$$
P(X_n = 0 text eventually) = sum_k=2^inftyP(tau_0 =k)
$$
where $tau_0 = infi ge 1 : X_i = 0 $.

Hint 3:
$P(tau_0 =k) = 0$ whenever $k$ is odd, and when it's not zero, it has a special form.